Problem: Luis is packing his bags for his vacation. He has $7$ unique Fabergé eggs, but only $3$ fit in his bag. How many different groups of $3$ Fabergé eggs can he take?
Answer: Luis has $3$ spaces for his Fabergé eggs, so let's fill them one by one. At first, Luis has $7$ choices for what to put in the first space. For the second space, he only has $6$ Fabergé eggs left, so there are only $6$ choices of what to put in the second space. So far, it seems like there are $7 \cdot 6 = 42$ different unique choices Luis could have made to fill the first two spaces in his bag. But that's not quite right. Why? Because if he picked Fabergé egg number 3, then Fabergé egg number 1, that's the same situation as picking number 1 and then number 3. They both end up in the same bag. So, if Luis keeps filling the spaces in his bag, making $7\cdot6\cdot5 = \dfrac{7!}{(7-3)!} = 210$ decisions altogether, we've overcounted a bunch of groups. How much have we overcounted? Well, for every group of $3$ , we've counted them as if the order we chose them in matters, when really it doesn't. So, the number of times we've overcounted each group is the number of ways to order $3$ things. There are $3! = 6$ ways of ordering $3$ things, so we've counted each group of $3$ Fabergé eggs $6$ times. So, we have to divide the number of ways we could have filled the bag in order by number of times we've overcounted our groups. $ \dfrac{7!}{(7 - 3)!} \cdot \dfrac{1}{3!}$ is the number of groups of Fabergé eggs Luis can bring. Another way to write this is $ \binom{7}{3} $, or $7$ choose $3$, which is $35$.